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(B) Let $r$ be the radius of the spherical balloon. Given that the volume $V = \frac{4}{3}\pi r^3$ is increasing at a rate of $\frac{dV}{dt} = 35 \, c

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$10$

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(B) Let $r$ be the radius of the spherical balloon. Given that the volume $V = \frac{4}{3}\pi r^3$ is increasing at a rate of $\frac{dV}{dt} = 35 \, cc/min$.We know that $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.Given diameter $d = 14 \, cm$,so radius $r = 7 \, cm$.Substituting the values: $35 = 4\pi (7)^2 \frac{dr}{dt} = 196\pi \frac{dr}{dt}$.Thus,$\frac{dr}{dt} = \frac{35}{196\pi} = \frac{5}{28\pi} \, cm/min$.The surface area $S = 4\pi r^2$. The rate of increase of surface area is $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.Substituting $r = 7$ and $\frac{dr}{dt} = \frac{5}{28\pi}$:$\frac{dS}{dt} = 8\pi (7) \left( \frac{5}{28\pi} ight) = 56\pi \left( \frac{5}{28\pi} ight) = 2 	imes 5 = 10 \, cm^2/min$.

$A$ spherical balloon is being inflated at the rate of $35 \, cc/min$. The rate of increase of the surface area of the balloon when its diameter is $14 \, cm$ is ....... $sq \, cm/min$.

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